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Relativistic Quantum Mechanics

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Relativistic Quantum Mechanics Relativistic Quantum Mechanics Dipankar Chakrabarti Department of Physics, Indian Institute of Technology Kanpur, Kanpur 208016, India (Dated: August 6, 2020) 1 I. INTRODUCTION Till now we have dealt with non-relativistic quantum mechanics. A free particle satisfying ~p2 Schrodinger equation has the non-relatistic energy E = 2m . Non-relativistc QM is applicable for particles with velocity much smaller than the velocity of light(v << c). But for relativis- tic particles, i.e. particles with velocity comparable to the velocity of light(e.g., electrons in atomic orbits), we need to use relativistic QM. For relativistic QM, we need to formulate a wave equation which is consistent with relativistic transformations(Lorentz transformations) of special theory of relativity. A characteristic feature of relativistic wave equations is that the spin of the particle is built into the theory from the beginning and cannot be added afterwards. (Schrodinger equation does not have any spin information, we need to separately add spin wave function.) it makes a particular relativistic equation applicable to a particular kind of particle (with a specific spin) i.e, a relativistic equation which describes scalar particle(spin=0) cannot be applied for a fermion(spin=1/2) or vector particle(spin=1). Before discussion relativistic QM, let us briefly summarise some features of special theory of relativity here. Specification of an instant of time t and a point ~r = (x, y, z) of ordinary space defines a point in the space-time. We’ll use the notation xµ = (x0, x1, x2, x3) ⇒ xµ = (x0, xi), x0 = ct, µ = 0, 1, 2, 3 and i = 1, 2, 3 x ≡ xµ is called a 4-vector, whereas ~r ≡ xi is a 3-vector(for 4-vector we don’t put the vector sign(→) on top of x. Consider two events in space-time (x0, x1, x2, x3) and (x0 + dx0, x1 + dx1, x2 + dx2, x3 + dx3) where x0 = ct so dx0 = cdt as c =velocity of light and is a constant. In three dimensional space we define the distance between two points. We generalize the notion of the distance between two points in space to the interval between two points in the space-time, say, ds. For ds to be same for all observer(ie, in all inertial frames), it must be invariant under Lorentz transformations and rotations. The interval is defined as 2 µ ν ds = gµνdx dx (1) 2 where gµν is the metric of the space-time. In Minkowski space 1 0 0 0 0 −1 0 0 g = . (2) µν 0 0 −1 0 0 0 0 −1 So, ds2 = (cdt)2 − ((dx1)2 + (dx2)2 + (dx3)2) = (cdt)2 − (dr~ )2 (3) µ 0µ µ ν µ Under Lorentz transformation x transforms as x = Λ νx where Λ ν is a 4 × 4 matrix representing the Lorentz transformation operator. For example, the operator for boost along x1 axis γ −γβ 0 0 µ −γβ γ 0 0 Λ = (4) ν 0 0 1 0 0 0 0 1 q where β = v/c and γ = 1/ 1 − (v/c)2. So, the transformed coordinates under the boost along x1: v v ct0 = γ(ct − x1), x01 = γ(x1 − ct), x02 = x2, x03 = x3. (5) c c Check that ds02 = (cdt0)2 − ((dx01)2 + (dx02)2 + (dx03)2) = γ2(cdt − βdx01)2 − γ2(dx1 − βcdt)2 − (dx2)2 − (dx3)2 = ds2 (6) i.e., ds2 is Lorentz invariant. ds2 can be both positive or negative unlike spatial distance (dr~ )2 which is always positive. If ds2 > 0 i.e., (cdt)2 > (dr~ )2, the interval is called "time-like" ds2 < 0 i.e., (cdt)2 < (dr~ )2, the interval is called "space-like" ds2 = 0 i.e., (cdt)2 = (dr~ )2, the interval is called "light-like". 3 covariant & contravariant vectors: Any quantity which transforms like xµ under Lorentz ∂ transformation is called a contravariant vector while anything which transforms like ∂xµ is called covariant vector. General convention for contravariant vector is aµ (i.e.,µ is in the superscript) ∂ and for covariant vector aµ (i.e, µ is in the subscript) i.e, ∂xµ = ∂µ. The inner product of a covariant vector and a contravariant vector is a Lorentz invariant(i.e., scalar). The contra and covariant vectors are related by X ν xµ = gµνx . (7) ν Using the convention of summation over repeated indices we can write the above eqn as xµ = ν ν µ µν gµνx where ν in gµν is repeated again in x and hence is summed over. Similarly, x = g xν. µν In Minkowski space, gµν = g . So, we have ν 0 1 2 3 0 0 x0 = g0νx = g00x + g01x + g02x + g03x = g00x = x (8) ν 0 1 2 3 1 1 x1 = g1νx = g10x + g11x + g12x + g13x = g11x = −x . (9) 2 3 Similarly x2 = −x and x3 = −x . Inner product or scalar product of two 4-vectors is defined as µ 0 1 2 3 0 0 1 1 2 2 3 3 A · B = A Bµ = (A B0 + A B1 + A B2 + A B3) = (A B − A B − A B − A B ) (10) 0 0 ~ ~ µ ν µν = A B − A · B = gµνA B = g AµBν. (11) Differential operators: ∂ 1 ∂ ∂ ∂ ∂ ∂ = = ( , , , ) (12) µ ∂xµ c ∂t ∂x1 ∂x2 ∂x3 1 ∂ = (∂ , ∂ , ∂ , ∂ ) = ( , ∇~ ) (13) 0 1 2 3 c ∂t 1 ∂ ∂µ = gµν∂ = ( , −∇~ ) (14) ν c ∂t The Lorentz invariant second order differential operator or the d’Alembertian operator is 1 ∂2 ∂2 ∂2 ∂2 1 ∂2 = ∂µ∂ = ( , −( + + )) = ( , −∇2). (15) µ c2 ∂t2 ∂x2 ∂y2 ∂z2 c2 ∂t2 2 2 We know the relativistic mass mr = γm and energy E = mrc = γmc . The energy-momentum 4-vector is pµ = (E/c, ~p) where ~p = γm~v. So, E (γmc2)2 p2 = g pµp = pµp = ( )2 − (~p)2 = − (γm~v)2 = m2c2 (16) µν ν µ c c2 4 (in the natural unit ~ = c = 1, p2 = m2). So, the relativistic energy momentum relation is given by E2 = (~p)2c2 + m2c4. Another useful quantity is µ p · x = p xµ = Et − ~p · ~x. (17) For non-relativistic particle (v << c), we can write q ~p2 E = ~p2c2 + m2c4 = mc2(1 + )1/2 (18) m2c2 ~p2 (~p)4 ~p2 = mc2(1 + − + ··· ) = mc2 + − · · · (19) 2m2c2 8m4c4 2m ~p2 Negelecting the higher oredr terms, the kinetic energy of a non-relativistic particle is 2m = E − mc2. A. Klein-Gordon Equation Schrodinger proposed a relavistic form of his non-relativistic equation (at the same time when he developed his non-relativistic(NR) equation). Klein and Gordon developed this equation at a later time and is knaown as Klein-Gordon(KG) equation. Schrodinger used the NR energy- p2 momentum dispersion relation E = 2m . Using the correspondence principle ∂ E → Eˆ = i , ~p → ~pˆ = −i ∇~ (20) ~∂t ~ p2 in Eφ(~r, t) = 2m φ(~r, t), we arrive at the Schrodinger equation for free particle. Now extend the same algorithm for relativistic particle with energy-momentum relation E2 = ~p2c2 + m2c4. So we get the relativistic wave equation E2φ(x) = (~p2c2 + m2c4)φ(x) (21) ∂2 ⇒ − 2 φ(x) = (− 2c2∇~ 2 + m2c4)φ(x) (22) ~ ∂t2 ~ 1 ∂2 m2c2 ⇒ − ∇~ 2 φ(x) = − φ(x) (23) c2 ∂t2 ~2 m2c2 ⇒ ( + )φ(x) = 0. (24) ~2 µ This equation is known as Klein-Gordon equation. Note that = ∂µ∂ is a Lorentz invariant quantity, so the KG equation is Lorentz invariant only if φ is Lorentz invariant or Lorentz 5 scalar. Thus KG equation describes the relativistic dynamics of a scalar particle. The plane wave solution of the KG eqn is φ(x) = Ne−i(Et−~p·~x) (25) √ where N is the normalization constant and energy E = ± ~p2c2 + m2c4 i.e., energy can be both positive and negative. Continuity Equation: Pre-multiply Eq.(23) by φ∗(x) to get 1 ∂2 m2c2 φ∗(x) − ∇~ 2 φ(x) = − φ∗(x)φ(x) (26) c2 ∂t2 ~2 Now take the complex conjugate of Eq.(23) and post-multiply with φ(x), which gives 1 ∂2 m2c2 ( φ∗)φ − (∇~ 2φ∗)φ = − φ∗(x)φ(x) (27) c2 ∂t2 ~2 Eq(26)-Eq(27) gives: 1 ∂2φ 1 ∂2φ∗ φ∗ − φ − (φ∗∇2φ − φ∇2φ∗) = 0 (28) c2 ∂t2 c2 ∂t2 1 ∂ i ∂φ ∂φ∗ ⇒ ~ φ∗ − φ + ∇~ · ~ φ∗∇~ φ − (∇~ φ∗)φ = 0 (29) c ∂t 2mc ∂t ∂t 2im 1 ∂ ⇒ ρ + ∇~ · ~j = 0 (30) c ∂t µ ⇒ ∂µj = 0 (31) This is the continuity equation for the KG eqn, where i ∂φ ∂φ∗ j0 = ρ = ~ φ∗ − φ (32) 2mc ∂t ∂t ~j = ~ φ∗∇~ φ − (∇~ φ∗)φ . (33) 2im Recall the continuity eqn for Schrodinger equation, ρ is the probability density and ~j is the probability current. Continuity equation has the interpretation of conservation of probability. It tells that if the probability of finding a particle in some region decreases, the probability of finding it out side that region increases, i.e., there is a flow of probability current so that the total probability remains conserved. Since the KG eqn also satisfies the same continuity eqn, it is natural to interpret ρ as the probability density and ~j as the probability current. [Note: Density 6 transforms like the 0th component of a 4−vector (jµ) under Lorentz transformation. Since φ is a Lorentz invariant quantity, φ2 does not transform like a density, but ρ defined in Eq.(32) does.] The probability density corresponding to the plane wave solution reads ρ = 2|N|2E. There are two major problems with the KG equation. (1) The eqn has both positive and negative energy solutions. The negative energy solution poses a problem! For large |~p| we can have large negative energy, i.e., the system become unbounded from below.
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